Using continuous compounding, we have:
90000=65452 x e^.091t
1.3750534743017784024934303000672=e^.091t
ln 1.3750534743017784024934303000672=ln e^.091t=.091t ln e=0.091t
t=3.5 years
Isn't this a subtraction problem, not a multiplication problem?
If John starts out with 20 fish and lets 6 go, he still has (20-6), or 14, fish.
5*5 = 25
5*9= 45
5*1= 5
25 dollars now add the cents
now it is $25.50
54044.55 ÷ 100 = 540.4455
The radius of tire is larger than radius of wheel by 5 inch
Step-by-step explanation:
We know that circumference of the circle is 2πr where “r” is the radius of the circle
Circumference refers to the dimension of the periphery of the circle. Since tires are put on the periphery of the wheel hence, we considered the circumferential aspect of the wheel.
Given-
Circumference of tires= 28π inches
2πr= 28π cancelling the common term “π” both sides
r (radius of the tires) = 14 inches
Circumference of the wheel rims= 18π
2πr= 18π cancelling the common term “π” both sides
r (radius of the tires) = 9 inches
Difference between the radius= 14-9= 5 inches
Hence, the difference between the radius of tires and the radius of the wheels is 5 inches
